1009. Product of Polynomials (25)

最新推荐文章于 2024-03-04 18:04:28 发布

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最新推荐文章于 2024-03-04 18:04:28 发布

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本文详细阐述了如何通过输入两个多项式的系数和指数来计算它们的乘积，并提供了具体的C++代码实现。

This time, you are supposed to find A*B where A and B are two polynomials.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 a_N1 N2 a_N2 ... NK a_NK, where K is the number of nonzero terms in the polynomial, Ni and a_Ni (i=1, 2, ..., K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10, 0 <= NK < ... < N2 < N1 <=1000.

Output Specification:

For each test case you should output the product of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate up to 1 decimal place.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 3 3.6 2 6.0 1 1.6

#include<iostream>
#include<vector>
#include<utility>
#include<iomanip>
using namespace std;

#define max 10001
int main()
{
	int n1,n2,i,j,x;
	float tmp;
	pair<int,float> a[max],b[max],ans[max];
	float c[2*max]={0.0};
	cin>>n1;
	pair<int,float> pair1;
	for(i=0;i<n1;i++)
	{
		cin>>x>>tmp;
		pair1.first=x;
		pair1.second=tmp;
		a[i]=pair1;
	}
	cin>>n2;
	for(i=0;i<n2;i++)
	{
		cin>>x>>tmp;
		pair1.first=x;
		pair1.second=tmp;
		b[i]=pair1;
	}
	for(i=0;i<n1;i++)
	{
		for(j=0;j<n2;j++)
		{
			pair1.first=a[i].first+b[j].first;
			pair1.second=a[i].second*b[j].second;
			c[pair1.first]+=pair1.second;
		}	
	}

	int cnt=0;
	for(i=0;i<2*max;i++)
	{
		if(c[i]!=0)
		{
			ans[cnt].first=i;
			ans[cnt].second=c[i];
			cnt++;
			
		}
	}
	cout<<cnt;
	for(i=cnt-1;i>=0;i--)
	{
		
			cout<<setiosflags(ios::fixed);
			cout.precision(1);
			cout<<" "<<ans[i].first<<" "<<ans[i].second;
	}
	return 0;
}